Question

A makes a bet with $B$ of $5s$. to $2s$. that in a single throw with two dice he :will throw seven before $B$ throws four. Each has a pair of dice and they throw simultaneously until one of them wins : find $B^{\prime }s$ expectation.

Answer 1

$\Rightarrow$Seven can thrown in six ways
$(5,2),(4,3),(3,4),(2,5),(1,6),(6,1)$
Probability of A throwing $\left(7\right)=\frac{6}{36}=\frac{1}{6}$
$\Rightarrow$Similarly four can be obtained in $3$ ways $(1,3),(3,1),(2,3)$
Probability of B throwing $\left(4\right)=\frac{3}{36}=\frac{1}{12}$
Probability of A winning$=\frac{1}{6}+\frac{5}{6}\times \frac{11}{12}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^2{\left(\frac{11}{12}\right)}^2\frac{1}{6}$
$=\frac{1}{6}\times \frac{1}{1-\frac{55}{72}}=\frac{12}{17}$
Probability of B winning$=\frac{5}{17}$
A gets $\frac{12}{17}\times 2\left(\frac{24}{17}\right)$ if he wins and pays $5\times \frac{5}{17}\left(\frac{25}{17}\right)$ to B if loss
B's expectation$=Rs(\frac{25}{17}-\frac{24}{17})$
$=\frac{1}{17}$